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Special Factoring: Factoring Sums and Differences
     of Cubes & Recognizing Perfect Squares (page 2 of 3)

Sections: Differences of squares, Sums and differences of cubes, Recognizing patterns

The other two special factoring formulas are two sides of the same coin: the sum and difference of cubes. These are the formulas:

a³ + b³ = (a + b)(a² – ab + b²)
a³ – b³ = (a – b)(a² + ab + b²)

You'll learn in more advanced classes how they came up with these formulas. For now, just memorize them. First, notice that the terms in each factorization are the same; then notice that each formula has only one "minus" sign. For the difference of cubes, the "minus" sign goes with the linear factor, a – b; for the sum of cubes, the "minus" sign goes in the quadratic factor, a² – ab + b². Some people use the mnemonic "SOAP" for the signs; the letters stand for "same" as the sign in the middle of the original expression, "opposite" sign, and "always positive".

a³ ± b³ = (a [same sign] b)(a² [opposite sign] ab [always positive] b²)

Whatever method helps you best keep these formulas straight, do it, because you should not assume that you'll be given these formulas on the test. You really should know them. Note: The quadratic part of each cube formula does not factor, so don't attempt it.

When you have a pair of cubes, carefully apply the appropriate rule. By "carefully", I mean "using parentheses to keep track of everything, especially the negative signs". Here are some typical problems: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

• Factor x³ – 8

This is x³ – 2³, so I get:

x³ – 8 = x³ – 2³
           = (x – 2)(x² + 2x + 2²)
        = (x – 2)(x² + 2x + 4)

• Factor 27x³ + 1

Remember that 1 can be regarded as having been raised to any power you like, so this is really (3x)³ + 1³.  Then I get:

27x³ + 1 = (3x)³ + 1³
               = (3x + 1)((3x)² – (3x)(1) + 1²)
               = (3x + 1)(9x² – 3x + 1)

• Factor x³y⁶ – 64

This is (xy²)³ – 4³, so I get:

x³y⁶ – 64 = (xy²)³ – 4³
                = (xy² – 4)((xy²)² + (xy²)(4) + 4²)
                = (xy² – 4)(x²y⁴ + 4xy² + 16)

There is one "special" factoring that can actually be done using the usual methods for factoring, but, for whatever reason, many texts and instructors make a big deal of treating this case separately. "Perfect square trinomials" are quadratics that you got by squaring a binomial. For instance, (x + 3)² = (x + 3)(x + 3) = x² + 6x + 9 is a perfect square trinomial.

Recognizing the pattern to perfect squares isn't a make-or-break issue, but it can be a time-saver occasionally. The trick is really quite simple: If the first and third terms are squares, figure out what they're squares of. Multiply those things, multiply that product by 2, and compare your result with the quadratic's middle term. If you've got a match, then you've got a perfect square.

• Is x² + 10x + 25 a perfect square trinomial? If so, write the trinomial as the square of a binomial.

Well, the first term, x², is the square of x. The third term, 25, is the square of 5. Multiplying, I get 5x. Multiplying this by 2, I get 10x. And this matches the middle term. So:

this quadratic is a perfect square: x² + 10x + 25 = (x + 5)²

• Write 16x² – 48x + 36 as a squared binomial.

The first term, 16x², is the square of 4x, and the last term, 36, is the square of 6. Actually, since the middle term has a "minus" sign, the 36 is the square of –6. Just to be sure, I'll make sure that the middle term matches the pattern: (4x)(–6)(2) = –48x. It's a match, so this is a perfect square:

16x² – 48x + 36 = (4x – 6)²

• Is 4x² – 25x + 36 a perfect square trinomial?

The first term, 4x², is the square of 2x, and the last term, 36, is the square of 6 (or, in this case, –6, if this is a perfect square). Checking the middle term, I get (2x)(–6)(2) = –24x, which does not match the middle term. So:

this is not a perfect square trinomial.

That's all there is to perfect squares.

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